Belorusija 2012

Solution:

Consider arbitrary arrangement of the boys along the circle. We put the signs "+" or "-" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign "+" before the boy if he is taller than the previous boy and we put the sign "-" if he is shorter than the previous one (see the fig.). It is evident that the boy is tall if and only if the sign "+" stands before him and the sign "-" stands after him; the boy is short if and only if the sign "-" stands before him and the sign "+" stands after him. (Note that since the tallest boy among all $N$ boys is tall, there exist the signs "+" as well as the signs "-" in any arrangement.) Therefore, the number of tall boys in the arrangement is equal to the number of alternations of "+" and "-"; the number of short boys in the arrangement is equal to the number of alternations of "-" and "+". Regarding all successive signs "+" as one sign "+" and all successive signs "-" as one sign "-", we construct a new arrangement of signs. It is easy to see that the number of alternations of "+" and "-" (of "-" and "+") in the initial arrangement is equal to the number of alternations of "+" and "-" (of "-" and "+") in the new arrangement. But it is evident that the number of alternations of "+" and "-" is equal to the number of alternations of "-" and "+" in the new arrangement. Therefore, the numbers of tall and short boys in any arrangement are equal.